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%I #4 May 13 2019 01:10:52
%S 1,1,1,2,1,2,1,2,1,1,1,2,1,1,2,1,1,1,2,1,1,2,1,1,1,2,1,1,2,1,2,1,2,1,
%T 1,1,2,1,1,2,1,2,1,1,1,2,1,1,2,1,2,1,2,1,1,1,2,1,1,2,1,2,1,1,1,2,1,1,
%U 2,1,2,1,2,1,1,1,2,1,1,2,1,2,1,1,1,2
%N Difference sequence of A325597.
%C All the terms of A325599 are in {1,2}, so that A325600 and A325601 are a complementary pair.
%H Clark Kimberling, <a href="/A325599/b325599.txt">Table of n, a(n) for n = 1..10000</a>
%t mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]); a = {1}; Do[AppendTo[a, mex[Rest[2 a] + Most[a], Last[a] + 1]], {60}]; a (* A325597 *)
%t c = Complement[Range[Last[a]], a] (* A325598 *)
%t da = Differences[a] (* A325599 *)
%t Flatten[Position[da, 1]] (* A325600 *)
%t Flatten[Position[da, 2]] (* A325601) *)
%t (* _Peter J. C. Moses_, May 07 2019 *)
%Y Cf. A325597.
%K nonn,easy
%O 1,4
%A _Clark Kimberling_, May 11 2019
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